The r.m.s speed of oxygen molecules at 47^circtextC$47^{\circ}\text{C}$ is equal to that of the hydrogen molecules kept at ________ ^circtextC$^{\circ}\text{C}$. (Mass of oxygen molecule / mass of hydrogen molecule = 32 / 2)

### Related Formula V_textrms = sqrtfrac3RTM$$V_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$ ### Core Logic Given that the RMS speed of oxygen molecules equals the RMS speed of hydrogen molecules: V_textrms(O_2) = V_textrms(H_2)$$V_{\text{rms}(O_2)} = V_{\text{rms}(H_2)}$$ sqrtfrac3RT_O_2M_O_2 = sqrtfrac3RT_H_2M_H_2$$\sqrt{\frac{3RT_{O_2}}{M_{O_2}}} = \sqrt{\frac{3RT_{H_2}}{M_{H_2}}}$$ Squaring both sides: fracT_O_2M_O_2 = fracT_H_2M_H_2$$\frac{T_{O_2}}{M_{O_2}} = \frac{T_{H_2}}{M_{H_2}}$$ ### Step 1: Absolute Temperature Conversion Temperature of oxygen in Kelvin: T_O_2 = 273 + 47 = 320 text K$$T_{O_2} = 273 + 47 = 320 \text{ K}$$ ### Step 2: Evaluating for Hydrogen Substitute the given mass ratio and absolute temperature: frac32032 = fracT_H_22$$\frac{320}{32} = \frac{T_{H_2}}{2}$$ 10 = fracT_H_22$$10 = \frac{T_{H_2}}{2}$$ T_H_2 = 20 text K$$T_{H_2} = 20 \text{ K}$$ ### Step 3: Final Conclusion Convert back to Celsius: T_H_2 text in ^circtextC = 20 - 273 = -253^circtextC$$T_{H_2} \text{ in }^{\circ}\text{C} = 20 - 273 = -253^{\circ}\text{C}$$ ### Pattern Recognition Direct proportionality between Temperature and Molecular Mass for equal RMS speeds. T_1/M_1 = T_2/M_2$T_1/M_1 = T_2/M_2$. Always convert to Kelvin before evaluating. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases